Relationship: reflexive, symmetric, antisymmetric, transitive

nicnicman
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Homework Statement


Determine which binary relations are true, reflexive, symmetric, antisymmetric, and/or transitive.

The relation R on P = {a, b, c} where R = {(a, a), (a, b), (a, c), (b, c), (c, b)}


Homework Equations





The Attempt at a Solution


Not reflexive because there is no (b, b) or (c, c).
Not symmetric because there is (a, b), but not (b, a).
Not antisymmetric because there is (b, c) and (c, b).
Not transitive because there is (b, c) and (c, b) but no (b, b).

Is it possible for there to not be any binary relations?
 
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nicnicman said:

Homework Statement


Determine which binary relations are true, reflexive, symmetric, antisymmetric, and/or transitive.

The relation R on P = {a, b, c} where R = {(a, a), (a, b), (a, c), (b, c), (c, b)}


Homework Equations





The Attempt at a Solution


Not reflexive because there is no (b, b) or (c, c).
Not symmetric because there is (a, b), but not (b, a).
Not antisymmetric because there is (b, c) and (c, b).
Not transitive because there is (b, c) and (c, b) but no (b, b).

Is it possible for there to not be any binary relations?

It is a binary relation, but as you say, it doesn't have any of those properties.
 
Yeah, that's what I thought. Thanks for the help again.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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